Students `J_(1)`, `J_(2)`, `J_(3)` and `J_(4)` perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum, they use different lengths of the pendulum and record time for different number of oscillations. The observations are shown in the table. Least count for length `= 0.1 cm`, least count for time = 1s
`{:("Students",underset("pendulum (cm)")("Length of the"),underset("oscillations(n)")("No.of"),underset("of pendulum (s)")("Time period")),(I_(1),100.0,20,20),(J_(1),400.0,10,40),(J_(1),100.0,10,20),(I_(1),400.0,20,40):}`
If `P_(1),P_(2),P_(3)` and `P_(4)` are the % error in g for students `J_(1)`, `J_(2)`, `J_(3)` and `J_(4)` respectively then -

Options

A. `P_(1) = P_(3)`
B. `P_(3)` is maximum
C. `P_(4)` is minimum
D. `P_(a) = P_(4)`

Explanation

To find the percentage error in the acceleration due to gravity (g) for students J1, J2, J3, and J4, we will follow these steps:

### Step 1: Understand the relationship between g, L, and T
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
From this, we can derive that:
\[ g = \frac{4\pi^2 L}{T^2} \]
This indicates that \( g \) is proportional to \( \frac{L}{T^2} \).

### Step 2: Determine the formula for percentage error in g
The percentage error in \( g \) can be expressed as:
\[ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} \]
Where:
- \( \Delta g \) is the absolute error in g,
- \( \Delta L \) is the absolute error in length,
- \( \Delta T \) is the absolute error in time.

### Step 3: Calculate the percentage error for each student
Given:
- Least count for length \( \Delta L = 0.1 \, \text{cm} = 0.001 \, \text{m} \)
- Least count for time \( \Delta T = 1 \, \text{s} \)

Now we will calculate \( P_1, P_2, P_3, \) and \( P_4 \) for students J1, J2, J3, and J4 respectively.

#### For Student J1:
- Length \( L = 100.0 \, \text{cm} = 1.0 \, \text{m} \)
- Total time for 20 oscillations \( T = 20 \, \text{s} \)
- Time for one oscillation \( t = \frac{20}{20} = 1 \, \text{s} \)

Calculating \( P_1 \):
\[ P_1 = \left( \frac{0.1}{100} + 2 \cdot \frac{1}{20} \right) \times 100 = \left( 0.001 + 0.1 \right) \times 100 = 0.6\% \]

#### For Student J2:
- Length \( L = 400.0 \, \text{cm} = 4.0 \, \text{m} \)
- Total time for 10 oscillations \( T = 40 \, \text{s} \)
- Time for one oscillation \( t = \frac{40}{10} = 4 \, \text{s} \)

Calculating \( P_2 \):
\[ P_2 = \left( \frac{0.1}{400} + 2 \cdot \frac{4}{40} \right) \times 100 = \left( 0.00025 + 0.2 \right) \times 100 = 0.42\% \]

#### For Student J3:
- Length \( L = 100.0 \, \text{cm} = 1.0 \, \text{m} \)
- Total time for 10 oscillations \( T = 20 \, \text{s} \)
- Time for one oscillation \( t = \frac{20}{10} = 2 \, \text{s} \)

Calculating \( P_3 \):
\[ P_3 = \left( \frac{0.1}{100} + 2 \cdot \frac{2}{20} \right) \times 100 = \left( 0.001 + 0.2 \right) \times 100 = 1.1\% \]

#### For Student J4:
- Length \( L = 400.0 \, \text{cm} = 4.0 \, \text{m} \)
- Total time for 20 oscillations \( T = 40 \, \text{s} \)
- Time for one oscillation \( t = \frac{40}{20} = 2 \, \text{s} \)

Calculating \( P_4 \):
\[ P_4 = \left( \frac{0.1}{400} + 2 \cdot \frac{2}{40} \right) \times 100 = \left( 0.00025 + 0.1 \right) \times 100 = 0.28\% \]

### Step 4: Summarize the results
- \( P_1 = 0.6\% \)
- \( P_2 = 0.42\% \)
- \( P_3 = 1.1\% \)
- \( P_4 = 0.28\% \)

### Conclusion
From the calculations, we see that the maximum percentage error is for student J3, and the minimum is for student J4. Thus, the correct options regarding the percentage errors are those that reflect this finding.